Rotational Motion

Definitions

Most of our discussion will concerned with rigid bodies— objects with definite shapes that don’t change

Purely rotational motion: all points in a body move in circle

r ℓθ

P

O

Radians

One radian is the angle created (subtended) by an arc whose length is equal to the radius

360o = 2 π rad

r ℓθ

P

O

Angular Velocity and Acceleration

Average angular velocity:

ω = Δθ / Δt Measured in radians per second

Average angular acceleration:

α = (ω – ω0)/ Δt = Δω / Δt Measured in radians per second squared

The Velocity of a Point

A point on a rotating wheel has the following linear velocity

v = Δℓ / Δt = r (Δθ/ Δt)

or

v = rω

Different Points Can Have Different Velocities

Despite the fact that ω is the same for all points, points with different values of r have different velocities

Tangential Velocities Can Be Different for Points on an Object

Tangential Velocities Can Be Different for Points on an Object

Acceleration

Angular acceleration is related to tangential linear acceleration by:

atan = Δv / Δt = r (Δω / Δt)or

atan = rα Total linear acceleration is:

a = atan + aR, Where aR is the radial or centripetal acceleration

toward the center of the object’s path

Centripetal Acceleration

aR = v2/ r = (ωr)2 / r = ω2r

Frequency and Period

Frequency is the number of complete revolutions (rev) per second

One revolution corresponds to an angle of 2π radians Therefore, 1 rev/sec = 2π rad/ sec

f = ω/ 2π or ω = 2πf The unit of frequency is the hertz (Hz)

1 Hz = 1 rev/s The time required for one revolution is a period T

T = 1/f

Centripetal Force

A force is required to keep an object moving in a circle If the speed is

constant the force is directed towards the center of the circle

∑FR = maR = mv2/r

There Is No Outward Force!

Components of Circular Motion

Force On a Revolving Ball (Horizontal)

Estimate the force a person must exert on a string attached to a 0.15 kg ball to make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second

F = mv2/r = (0.15 kg)(7.54 m/s)2/(0.6 m)

≈ 14 N This solution ignores the fact that the ball cannot

be perfectly horizontal because it has weight due to the force of gravity

Force On a Revolving Ball (Vertical)

A 0.15 kg ball on the end of a 1.1 m string is swung in a vertical circle. Determine the minimum speed the ball must have at the top of its arc so that it continues moving in a circle. Calculate the tension in the string at the bottom of the arc if the ball is moving at twice the minimum speed.

Solution At the top of the arc there

are two forces on the ball: mg and the tension FTA

∑FR = maR

FTA + mg = mvA2/r

The minimum speed occurs when FTA = 0

mg = mvA2/r

vA= √(gr)

= 3.28 m/s

Solution (cont’d)

At the bottom of the circle the cord exerts its tension force FTB upward, but the force of gravity mg is downward

∑FR = maR

FTB – mg = mvB2/r

= mvB2/r + mg

For v = 6.56 m/s (2x minimum) F= 7.34 N

Ferris Wheel

A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v

Is the normal force the seat exerts on the rider at the top of the circle less than, more than, or the same as the force exerted at the bottom of the arc?

Solution

This is exactly like the vertical string problem with FN replacing tension. Therefore, the force at the top is less than the force at the bottom

Centripetal Forces--Uniform Circular Motion

Roller Coaster Physics

Demonstrations

A Right Hand Turn

Forces on Cars In Turns

Flat Road Banked Turn

Centripetal acceleration is horizontal, not parallel to road surface

Banked vs Flat Turns

Circular Motion as the Limit of Straight Line Motion

Ball on a String

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c8_whirligig.html

Gravitron

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c7_rotor.html

Objects on a Turntable

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c7_turntable.html

Vertical Circular Motion

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c8_vertical.html

Kinematic Equations

Angular

ω = ω0 + ½αt

θ = ω0t + ½αt2

ω2 = ω02 + 2αθ

ω = (ω + ω0)/2

Linear

v = v0 + at

x = v0t + ½ at2

v2 = v02 + 2ax

v = (v + v0)/2

Example

A bicycle slows down from 8.4 m/s to rest over a distance of 115 m. Each wheel has a diameter of 68.0 cm.

Determine the angular velocity of the wheels at the initial moment; the total number of revolutions each wheel makes in coming to rest, and the time it took to stop.

Solution

At the initial instant, points on the rim of the wheel are moving at 8.4 m/s.

The initial angular velocity is:

ω0 = v0/r = (8.4 m/s)/ (0.34 m) = 24.7 rad/s 115m of ground passes under the bike as it

stops. Each revolution of a wheel corresponds to a distance of 2πr so:

115 m/ 2πr = 115 m/(2πr)(0.34 m) = 53.8 rev

Solution

Angular acceleration of the wheel can be obtained from:

ω2 = ω02 + 2αθ

α = (ω2 - ω02)/ 2θ = 0 – (24.7 rad/s)2

2(2π)(53.8 rev)= -0.902 rad/s2

From ω = ω0 + ½αt we get that:

t = (ω - ω0)/α = (0 – 24.7 rad/s)/ -0.902 rad/s2 = 27.4 s

Rolling

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c15_rolling.html

Rotational Dynamics

Up to this point we have been studying rotational kinematics— how things move

Now we will go on to rotational dynamics—why things move

Ferris Wheel Kinematics

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c13_consta_ex.html

Torque

Causing an object to rotate around its axis requires a force The effect of a force is greater if it is placed further from the

axis of rotation

The angular acceleration of an object is directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts This distance is called a lever arm

The product of the force times the lever arm is called torque (α proportional to τ)

Only the Perpendicular Component of Force Contributes to Torque

Torque = r┴F = rF┴

Example

The biceps muscle exerts a vertical force on the lower arm. Calculate the torque about the axis of rotation through the elbow joint assuming the muscle is attached 5.0 cm from the elbow

Do Now (10/1/13):

The biceps muscle exerts a vertical force on the lower arm. Calculate the torque about the axis of rotation through the elbow joint assuming the muscle is attached 5.0 cm from the elbow

Solution

F = 700 N and r┴ = 0.05m so

τ = r ┴F = (0.05 m)(700 N) = 35 m-N

Solution

r┴ = (0.05 m)(sin 60o)

Therefore

τ = (0.05 m)(sin 60o)(700N)

= (0.05 m)(0.866)(700N)

= 30 m-N

Torque on a Compound Wheel

Two thin cylindrical wheels of radii r1 = 30 cm and r2 = 50 cm are attached to each other as shown. Calculate the net torque on this wheel due to the two forces shown, each of magnitude 50 N

Solution

The two forces create torques in different directions We can consider one to be positive an done to be

negative Because F2 is not perpendicular to the axis of

rotation we must only use the component of the force that is perpendicular

τ = r1F1 – r2F2sin60o

= (0.3m)(50N) – (0.5m)(50 N)(0.866)= -6.7 m-N

Balancing Torques

Force = 2FForce = F

X X

AB C

D

A force of magnitude F is applied at a distance X from the center of a seesaw. Another force of magnitude 2F is also applied to the seesaw at a distance X on the other side of the fulcrum. At what location(s) and in what directions can a third force of magnitude F be applied so that the seesaw is balanced?

Solution

Force = 2FForce = F

X X

AB C

D

Force = -F

Force = F

The seesaw can be balanced if a force of magnitude F is applied in the positive direction at A or in the negative direction at D

Torques

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c14_equilibrium.html

Rotational Analogs of Mass and Momentum

The angular acceleration of an object is directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts α is proportional to Στ

Rotational Analog of Mass

Consider a point of mass m on the end of a string with an applied force F

From F = ma we draw the analog that= mrα

Multiplying both sides by r we find that Τ = rF = mr2α

The quantity mr2 is called the moment of inertia of the mass (I)

The moment of inertia is the rotational analog of mass

r

F

Rotational Analog of Newton’s Second Law

From the previous page

τ = rF = mr2α Therefore

Στ = Σ (mr2)α But mr2 = I, therefore

Στ = Iα

Question

An ice skater in a spin turns rapidly if her arms and legs are in line with her body, but turns more slowly when they are outstretched. Why?

The moment of inertia (rotational analog of mass) is mr2. When more mass is held at greater distance from the body, I is increased.

Rotational Analog of Momentum

Define angular momentum asL = Iω

The total angular momentum of a rotating body remains constant if the net torque acting on it is zero (∆L/∆t = 0)

Angular momentum is a vector quantity We use the right hand rule to determine the

direction of L

Right Hand Rule

Numerous physical phenomena have resultant vectors that are perpendicular to the original vectors in the problem

In these cases a standard convention is needed to establish positive or negative vector direction

The “right hand rule” is used in these cases

Conservation of Angular Momentum

Consider a man walking on a circular platform that is at rest

If the person starts walking forward the platform will turn backward

Newton’s 3rd Law provides a rationale for this but so does conservation of angular momentum

• Man’s angular momentum = Iω = (mr2)(v/r) where v is his velocity relative to the earth, r is his distance from the axis and m his mass

• The total angular momentum of the man-platform system remains zero

Classwork

Problem 43 in textbook ch. 8 Use the rest of class to work on hw

Conservation of Momentum

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c16_collision1.html

Angular Momentum

http://webphysics.davidson.edu/physlet_resources/bu_semester1/c15_race.html

Do Now (10/2/13):

1. If you walked along a tall fence, why would holding your arms out help you to balance?

2. Consider two rotating bicycle wheels, one filled with air and the other filled with water. Which would be more difficult to stop rotating? Explain.

Conceptual Questions

Work with your group to answer the review questions

Submit one paper per group Each group member must write answers!!!

Use different color inks/pencils for different group members

*Bonus – answer the “Think and Explain” questions

Bicycle Wheel Gyroscope

A rapidly spinning wheel has angular momentum. Changing the direction of the wheel will require an applied torque or will result in a torque to conserve angular momentum.

Initial angular momentum of man-gyro system is +L

After wheel is turned over the momentum of the wheel is –L

Therefore, the man-wheel system will have to have an angular momentum of +2L upward

Do Now (10/3/13):

You sit in the middle of a large, freely rotating turntable at an amusement park. If you crawled toward the outer rim, what would happen to the rotation speed? Explain.

How do clockwise and counterclockwise torques compare when a system is balanced?

Gravity

Universal Gravitation

Every object attracts every other object with a force that for any two objects is directly proportional to the mass of each object.

F = G m1 m2

d2

G = 6.67 x 10 –11 N m2/kg2

Gravity is a very weak force!! It is the weakest of the four fundamental forces

Electromagnetic force, nuclear strong force, nuclear weak force.

The Inverse-Square Law

The force of gravity decreases with the square of the distance between objects The force of gravity reduces rapidly with

distance. However, the force of gravity never reaches

zero.

Gravitational Fields

We define the gravitational acceleration “g” as g = F/m

For an object on Earth’s surface, it is at a radius R from the center.

Therefore,g = F/m = (G m M/R2)/m

= GM/R2 = 9.8m/s2

Where M is the mass of the Earth (5.98 x 1024

kg) and R = 6.37 x 106 km

Attraction Between People

What is the attraction between a 50 kg person and a 75 kg person?

F = (6.67 x 10-11 Nm2/kg2)(50 kg)(75 kg)

(0.50 m)2

= 1 x 10-6 N (not very much force)

Gravity on Mt. Everest

Replace r (radius of earth) by r + 8.8 km = 6389 km

g = GmE/r2 = (6.67 x 10-11 Nm2/kg2)(5.98 x 1024)

(6.389 X 108 m)2

= 9.77 m/s2 compared with 9.8 m/s2

The Falling Moon

Newton compared the moon to a cannonball fired from the top of a high mountain. He imagined the mountaintop to be above the

atmosphere. He reasoned that a cannonball fired fast enough

would go into orbit around the Earth Essentially, Newton reasoned that the moon

was falling into the Earth

Orbiting the Earth

Geosynchronous Satellites

G (msatmE)/r2 = msat (v2/r) But v = 2πr/T where T = 1 day = 84,600 s Therefore,

G(mE/r2) = (2πr)2/rT2

r3 = 7.54 x 1022 m3

r = 4.23 x 107 m

r = 26,283 mi from center of earth

= 22,317 mi above surface

Kepler’s Laws

50 years before Newton Kepler had derived the Laws of Planetary Motion

1st law: Path of each planet about the sun is an ellipse

2nd Law: Each planet moves so that an imaginary line from the sun to the planet sweeps out equal areas in equal amounts of time

Kepler’s Law Animation

Kepler’s Laws

3rd Law: The ratio of the squares of the periods (time for one revolution about the sun) of any two planets is equal to the ratio of their mean distances cubed from the sun

(T1/T2)2 = (r1/r2)3

We can rewrite this as:

r13/T1

2 = r23/T2

2

Which is the same for all planets

Enter Newton

Gm1Ms/r12 = m1v1

2/r1

where Ms is the mass of the sun Since v1 = 2πr1/T1,

Gm1Ms/r12 = m1(4π2r1)/T1

2

Rearranging we get:T1

2/r13 = 4π2/GMs

Since we can derive this same expression for another planet m2, then we can set

r13/T1

2 = r23/T2

2

Using Kepler’s Laws

A Martian year is 1.88 Earth years. Determine the mean distance of Mars from the sun.

Let TE = 1 year and rES = 1.5 x 1011 m Kepler’s 3rd Law gives us:

rMS/rES = (TM/TE)2/3 = (1.88 yr/1 yr)2/3 = 1.53

times as far from the sun as the Earth

Geosynchronous Satellites Using Kepler’s Laws

The moon circles the Earth in roughly 27 days and rME = 380,000 km from earth

A geosynchronous satellite has a period of 1 day, therefore:

rSat = rME(Tsat/TM)2/3

= rME (1 day/27 day)2/3

= rME(1/3)2 = rME/9

Determining the Mass of the Sun

If rES = 1.5 x 1011 m determine the mass of the sun From a previous slide:

Gm1Ms/r12 = m1(4π2r1)/T1

2

MS = 4π2rES3/GTE

2 = 4π2(1.5 x 1011)/G(3.16 x 107)

= 2.0 x 1030 kg To calculate this we used:

TE = 1 yr = (365¼ d) (24 h/d) (3600 s/h)

= 3.16 x 107

Experiencing Weightlessness

The apparent weight of an object on earth is equal to:

w = mg + ma An object accelerating away from the

earth will appear to have increased weight An object falling towards earth will appear

to have reduced weight or weightlessness

Riding an Elevator

Weightlessness in Orbit

Since a satellite can be considered as “falling towards Earth”, people in a satellite experience “free fall” which provides the appearance of weightlessness.

This is different from the weightlessness experienced at distances far from massive objects

End

Do Now (10/4/13):

1. If the moon falls, why doesn’t it get closer to the earth?

2. Since the planets are pulled to the sun by gravity, why don’t they simply crash into the sun?

Practice:

Work on homework quietly!!!